# How do you find the definite integral of #int (1+3x)dx# from #[ -1,5]#?

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To find the definite integral of ( \int_{-1}^{5} (1+3x) , dx ), you first need to find the antiderivative of the function ( 1+3x ), which is ( x + \frac{3}{2}x^2 ). Then, you evaluate this antiderivative at the upper limit of integration (5) and subtract the value of the antiderivative evaluated at the lower limit of integration (-1).

So,

[ \int_{-1}^{5} (1+3x) , dx = \left[ x + \frac{3}{2}x^2 \right]_{-1}^{5} ]

[ = \left(5 + \frac{3}{2}(5)^2\right) - \left((-1) + \frac{3}{2}(-1)^2\right) ]

[ = \left(5 + \frac{3}{2}(25)\right) - \left((-1) + \frac{3}{2}(1)\right) ]

[ = \left(5 + \frac{75}{2}\right) - \left((-1) + \frac{3}{2}\right) ]

[ = \left(5 + \frac{75}{2}\right) - \left(-1 + \frac{3}{2}\right) ]

[ = \left(\frac{10}{2} + \frac{75}{2}\right) - \left(\frac{-2}{2} + \frac{3}{2}\right) ]

[ = \frac{85}{2} - \frac{1}{2} ]

[ = \frac{84}{2} ]

[ = 42 ]

Therefore, the definite integral of ( \int_{-1}^{5} (1+3x) , dx ) from -1 to 5 is 42.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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